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Electromagnetic Induction and Alternating Current

Magnetic  Flux

Magnetic_flux

 

The magnetic flux is proportional to the number of

field lines that pass through the surface.

 

Φ = Bn A

 

Φ   magnetic flux in Wb (Weber).  Also :  T.m2 of Vs

Bn   the component of B perpendicular to the  surface in T(esla)

A     area of the surface  in m2

 

Example

See the drawing above.

B = 1.5 x 10-3 T

A = 500 cm2

α = 40 o

 

Resolve B in both components  and calculate Bn

 

sin α = Bn/B 

Bn = B sin 40 o = (1.5 x 10-3 )( 0.643) = 9.642 x 10-4  T

 

Φ = Bn A

Φ = (9.642 x 10-4 )( 500 x 10-4 )= 4.82 x 10-5 Wb

 

Change of magnetic flux and electromagnetic induction                           

 A change in the magnetic flux in a closed circuit (coil) 

induces  a potential difference .

 

Law of Faraday:

 

Law_of_Faraday

 

 

 

                                                                                                                        

 

 

 

Vind  average induced potential difference in V

N      amount of loops of the coil

ΔΦ   change of magnetic flux in Wb

Δ t    time interval in which change occurs in s

 

Example

magnetic_induction

See the figure alongside.

The area of the turn is  40 cm2

The magnitude of the magnetic induction is

0.020 T

The turn is rotated.

At the moment of the figure the field lines are parallel to  the turn.

So the magnetic flux is 0 Wb

After  0.0050 s the turn has rotated  90o . In that  situation the magnetic flux is maximal

Determine the average induced potential difference (emf) in this time interval.

 

Vind =- N ( ΔΦ/Δt)

induction_current

 ΔΦ = Φmax – 0 = B A = (0.020 )(40 x 10-4) = 8.0 x 10-5 Wb

Vind = (1) (8.0 x 10-5)/0.0050 = 1.6 x 10-2 V

 

An induction current is produced when A is connected to B

The resistance of the circuit is 0.20 Ω

Uind = Iind R

1.6 x 10-2 = (Iind )(0,20)

Iind = 8,0.10-2 A

 

 

Sinusoidal alternating current

Peak voltage and root mean square value of voltage

 

V(t) = V0  sin (2πft)        (set calculator to radians)

 

V         voltage in V

V 0       peak voltage in V

f          frequency in Hz

t          time in s 

 

Vrms = ½ 2 V 0      

                                                                                                             

Vrms     root-mean-square value of           

Sinusoidal_alternating_current

             voltage

 

Example 1                                                           

 

 In the drawing aside:                              

V 0 =10 V

T = 50 ms

 f = 1/T=1/(50 x 10-3) = 20 Hz

So :

V(t) = 10 sin (2π 20 t)

 

Vrms = ½ 2. V0 = ½ 2. 10 = 7.0 V

 

This means :

A  DC voltage of  7.0 V produced the same power as an AC voltage  with a peak voltage of

10.0 V

 

Example 2

The mains voltage is :  V =230 V  (This means :  V rms = 230 V)

Find the peak voltage.

 

Vrms = ½ 2  V 0

 

230 = ½ 2  V 0

 

V 0 = 325 V 

 

 

Peak current and root-mean-square value of the current

In the previous paragraph we see : V(t) = V0  sin (2πft)       

 

Correspondingly  we can write for the current:

 

I(t) = I 0 sin (2πft)   

I rms = ½ 2. I0

 

Transformer

Transformer

 

VP  :  VS  = NP :  NS

 

VP   voltage across the primary coil in V

VS    voltage induced in the secondary coil in V

NP    amount of turns in the primary coil                                                     

NS    amount of turns in the secondary coil                                             

                                                                                                                     

 

PP = PS                                   (Assuming 100 % energy transfer, no energy dissipated in heating))

VP IP = VS IS

 

PP    power of the primary coil W (of VA)

IP     current in the primary coil in A

PS      power of the secondary coil  in W (of VA)

 IS       current  in the secondary coil  in A

 

Examples

1.         The primary coil of a transformer consists of 400 turns. The secondary coil  

      80 turns.

      The voltage across the primary coil is 230 V

      Find the voltage across the secondary coil

 

      VP  :  VS  = NP :  NS

      230 : VS = 400 : 80

      230 : VS = 5 : 1

      (5)(Vs) = (230) (1)

      VS = 230/5 = 46 V

 

2.           The primary coil is connected with an AC source

A bulb is connected with the secondary coil.

The voltage across the bulb is 6.0 V and

the current through the bulb is  0.4 A

The turns ratio is 5:1

 (NP : NS  = 5: 1)

 

Calculate the current in the primary coil

Primary_secondary_coil

        PS = VS IS = (6.0) (0.4) = 2.4 W

        VP = 5 VS = (5) (6.0) = 30 V

        PP = PS = 2.4 W

        PP = VP IP

             2.4 = (30)( IP)

             IP = 2.4/30 = 0.8 A